A body of mass 2kg moves under the influence of a force- Its position x changes with time according to the relation x-t3-3 where x is in meter and t in seconds- The work done by this force in first two seconds will be

x = t^3/3 , v = dx/dt = t^2 w = KE, v_1 = 0, v_2 = 4 w = 1/2mv_2^2 1/2mv_1^2 w = 1/2× 2 [(4)^2 (0)^2] = 16 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Physics

Relation between Work Done and PE

A body of mas...

Question

A body of mass 2kg moves under the influence of a force. Its position x changes with time according to the relation $x = \frac{t^{3}}{3}$ where x is in meter and t in seconds. The work done by this force in first two seconds will be

1600 Joule

No worries! We‘ve got your back. Try BYJU‘S free classes today!

160 Joule

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16 Joule

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.6 Joule

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 16 Joule
$x = \frac{t^{3}}{3}$ , $v = \frac{d x}{d t} = t^{2}$
$w = △ K E$ , $v_{1} = 0$ , $v_{2} = 4$
$w = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$
$w = \frac{1}{2} \times 2 [(4)^{2} - (0)^{2}] = 16$

Suggest Corrections

Similar questions

Q. A body of mass 2kg moves under the influence of a force. Its position x changes with time according to the relation

x = \frac{t^{3}}{3}

where x is in meter and t in seconds. The work done by this force in first two seconds will be

Q. Under the action of a force, a

2 k g

body moves such that its position

x

as a function of time is given by

x = t^{3} / 3

, where

x

is in meter and

t

in seconds. The work done by force in first two seconds is

Q. The relation between displacement

x

and time

t

for a body of mass

2 K g

moving under the action of a force is given by

x = \frac{t^{3}}{3}

, where

x

is in meter and

t

in second, calculate the work done by the body in first

2

seconds.

Q. Under the action of a force, a

2 kg

body moves such that its position

x

as a function of time t is given by

x = \frac{t^{3}}{3}

, where

x

is in meter and

t

in second. The work done by the force in the first two seconds is

Q. A body of mass

2 k g

moving under a force has relation between displacement

x

and time

t

x = \frac{t^{3}}{3}

where

x

is in metre and

t

is in sec. The work done by the body in first two second will be

Join BYJU'S Learning Program

A body of mass 2kg moves under the influence of a force. Its position x changes with time according to the relation x=t33 where x is in meter and t in seconds. The work done by this force in first two seconds will be

The correct option is C 16 Joule x=t33 , v=dxdt=t2 w=△KE, v1=0, v2=4 w=12mv22−12mv21 w=12×2[(4)2−(0)2]=16

A body of mass 2kg moves under the influence of a force. Its position x changes with time according to the relation $x = \frac{t^{3}}{3}$ where x is in meter and t in seconds. The work done by this force in first two seconds will be

The correct option is C 16 Joule
$x = \frac{t^{3}}{3}$ , $v = \frac{d x}{d t} = t^{2}$
$w = △ K E$ , $v_{1} = 0$ , $v_{2} = 4$
$w = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$
$w = \frac{1}{2} \times 2 [(4)^{2} - (0)^{2}] = 16$